Question

One half of an electrochemical cell consists of a pure nickle electrode in a 0.001 molal...

One half of an electrochemical cell consists of a pure nickle electrode in a 0.001 molal solution,

and the other half is a cadmium electrode immersed in a 0.5 molal solution. The electrode

potential is given by , where is the standard electrode potential and c is the molal concentration

of the ions in the solution. The cell potential is most likely?

a) -0.65 V

b) -0.15 V

c) -0.073 V

d) +0.75 V

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Standard electrode potential for Cd and Ni are

E0Cd2+/Cd = -0.403 V and E0Ni2+/Ni = -0.250 V

Hence Cd2+/Cd will be the anode and Ni2+/Ni will be the cathode

Using Nernst equations

E = E0 + (0.059/2) * log (Cion)

ECd2+/Cd = E0Cd2+/Cd + (0.059/2)log(0.5) =-0.403 + (0.059/2)log(0.5) = -0.4118 V

Similarly,

ENi2+/Ni = E0Ni2+/Ni + (0.059/2)log(0.5) =-0.250 + (0.059/2)log(0.001) = -0.3385 V

Cell potential will be

E = ENi2+/Ni - ECd2+/Cd = -0.3385 – (-0.4118) = + 0.0733 V

note: may be the option " c) -0.073" be replaced by the option " c) +0.073"

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A voltaic cell contains two half-cells. One half-cell contains a nickel electrode immersed in a 1.00...
A voltaic cell contains two half-cells. One half-cell contains a nickel electrode immersed in a 1.00 M Ni(NO3)2 solution. The second half-cell contains a titanium electrode immersed in a 1.00 MTi(NO3)3 solution. Ni2+(aq) + 2 e− → Ni(s)     E⁰red  = −0.257 V Ti3+(aq) + 3 e− → Ti(s)     E⁰red = −1.370 V (a) Using the standard reduction potentials given above, predict the standard cell potential of the voltaic cell. ____V (b) Write the overall balanced equation for the voltaic cell....
Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a...
Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a pair of Zn electrodes, each immersed in ZnSO4 solution. One solution is created to be exactly 0.500 M, and the potential on its electrode is measured to be 0.075 V higher than the other electrode. What is the concentration of the other solution? Enter your answer in M.
Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a...
Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a pair of Zn electrodes, each immersed in ZnSO4 solution. One solution is created to be exactly 0.400 M, and the potential on its electrode is measured to be 0.045 V higher than the other electrode. What is the concentration of the other solution? Enter your answer in M.​
From the information provided, use cell notation to describe the following systems: (a) In one half-cell,...
From the information provided, use cell notation to describe the following systems: (a) In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO3)2 solution with all solute concentrations 1 M. (b) The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution. (c) One half-cell consists of a silver electrode in a 1 M AgNO3...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.40 Mand 0.130 M , respectively. What is the initial cell potential? What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ? What is the concentrations of Ni2+when the cell potential falls to 0.46 V? What is the concentration of Zn2+when the cell potential falls to 0.46 V?
Use the standard reduction potentials shown here to answer the questions. Reduction half-reaction E∘ (V) Cu2+(aq)+2e−→Cu(s)...
Use the standard reduction potentials shown here to answer the questions. Reduction half-reaction E∘ (V) Cu2+(aq)+2e−→Cu(s) 0.337 2H+(aq)+2e−→H2(g) 0.000 A copper, Cu(s), electrode is immersed in a solution that is 1.00 M in ammonia, NH3, and 1.00 M in tetraamminecopper(II), [Cu(NH3)4]2+. If a standard hydrogen electrode is used as the cathode, the cell potential, Ecell, is found to be 0.073 V at 298 K. Part A Based on the cell potential, what is the concentration of Cu2+ in this solution?...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.10×10−2 M and 1.60 M , respectively. The initial potential is 0.51 V and the cell potential is 0.45 V when the concentration of Cu2+ has fallen to 0.230 M. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?
Electrochemical Cell Potentials Table 1: Electrochemical Cell Potentials Cell Measured Total Potential from Multimeter (V)1 Individual...
Electrochemical Cell Potentials Table 1: Electrochemical Cell Potentials Cell Measured Total Potential from Multimeter (V)1 Individual Half-Cell Potentials Cell Reactions5 ΔG (kJ)6 (Cu) Electrode Standard Potential (V)2 Metal Electrode Experimental Potential (V)3 Metal Electrode Theoretical Potential (V)4 Metal Electrode Potential % Error Cu | Sn 0.469 0.34 V 0.34-0.469 =-0.129 Cathode:       Cu2+ + 2e- -> Cu Anode:            Sn ->Sn2++ 2e- Net:          Cu2+ + Sn -> Cu + Sn2+ Cu | Al 0.796 0.34 V 0.34-0.796 =-0.456 Cathode:   Cu2+ +...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C ....
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C . The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.50 M , respectively. a) What is the initial cell potential? b) What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? c) What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ?
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.1×10−2 M and 1.400 M , respectively. A) What is the initial cell potential? ___V B)What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? ___V C)What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ? ___V
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT