Question

Consider a galvanic cell based upon the following half reactions: Ni2+ + 2e- → Ni -0.27...

Consider a galvanic cell based upon the following half reactions:

Ni2+ + 2e- → Ni -0.27 V

Cr3+ + 3e- → Cr -0.73 V

How many of the following responses are true?

1. Adding equal amounts of water to both half reaction vessels will decrease the potential of the cell

2. Increasing the mass of the Ni will change the initial potential of the cell

3. Cr is being oxidized during the reaction

4. Decreasing the concentration of Ni2+ (assuming no volume change) will decrease the potential of the cell

5. Decreasing the concentration of Cr3+ (assuming no volume change) will decrease the potential of the cell

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

3. Cr acts as anode. So, it will be oxidized. So, it is TRUE.

4. As Concentration of Ni2+ is in denominator. So, decreasing its concentration will decrease the concentration factor of Nernst equation. So, cell potential will decrease. Hence, this statement is TRUE.

5.Decrease in concentration of Cr3+ will increase the cell potential. Hence, this statement is FALSE.

So, correct statements are 1), 3) and 4).

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
Consider a galvanic cell based upon the following half reactions: Fe3+ + 3e- → Fe -0.0036...
Consider a galvanic cell based upon the following half reactions: Fe3+ + 3e- → Fe -0.0036 V Cu2+ + 2e- → Cu 0.34 V How many of the following responses are true? 1. Increasing the mass of the Cu will change the initial potential of the cell 2. Fe is being oxidized during the reaction 3. Increasing the concentration of Cu2+ (assuming no volume change) will decrease the potential of the cell 4. Decreasing the concentration of Fe3+ (assuming no...
calculate the potential of the cell: ni2+ +fe > ni+fe2+ the half reactions involved are: ni2+...
calculate the potential of the cell: ni2+ +fe > ni+fe2+ the half reactions involved are: ni2+ +2e- > ni E°=-0.26v and fe2+ +2e- > fe E°=-0.45v is the reaction spontaneous or nonspontaneous?
Consider a galvanic cell based on the following half reactions: E° (V) Al3+ + 3e- →...
Consider a galvanic cell based on the following half reactions: E° (V) Al3+ + 3e- → Al -1.66 Ba2+ + 2e- → Ba -2.90 If this cell is set up at 25°C with [Ba2+] = 2.50 × 10-3M and [Al3+] = 4.00 × 10-2M, the expected cell potential is ____ V
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 1.50molL−1 and 0.100 molL−1, respectively. Zn2+(aq)+2e−→Zn(s) E∘=−0.76V Ni2+(aq)+2e−→Ni(s) E∘=−0.23V What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.46 V?
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?
The following reactions take place in a galvanic cell: (i) Cu2+(aq)+ Ni (s)→ Cu(s) + Ni2+(aq)...
The following reactions take place in a galvanic cell: (i) Cu2+(aq)+ Ni (s)→ Cu(s) + Ni2+(aq) (ii) 2Ag+ (aq) +H2 (g) → 2Ag(s) +2H+ (aq) (iii) Cl2 (g) + Sn2+ (aq) → Sn4+ (aq) + 2Cl- (aq) (a) For each of the above spontaneous cell reactions, write the electrochemical cell using standard cell notation. (b) Use the standard reduction potentials below to evaluate EƟcell for the overall cell reaction taking place in (iii): Cl2(g) + 2e- → 2Cl- (aq) E...
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 M and 0.130 M , respectively. The volume of half-cells is the same. Part A: What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M ?  
A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell. The initial concentration...
A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell. The initial concentration of Ni2+(aq) in the Ni2+−Ni half-cell is [Ni2+]= 1.00×10−2 M . The initial cell voltage is +1.12 V . a. By using data in Table 20.1 in the textbook, calculate the standard emf of this voltaic cell. b. Will the concentration of Ni2+(aq) increase or decrease as the cell operates? c. What is the initial concentration of Ag+(aq) in the Ag+−Ag half-cell?
Calculate the cell potential for the voltaic cell based on the following half reactions at T...
Calculate the cell potential for the voltaic cell based on the following half reactions at T = 25ºC: Cr3+(aq) + 3e- → Cr(s) Eº = - 0.74 V TiO2+(aq) + 2H+ (aq) + 1e- → Ti3+(aq) + H2O(l) Eº = + 0.10 V Where, [Cr3+] = 1.0 x 10-4 M, [TiO2+] = 1.0 x 10-1 M, [H+] = 1.0 M, [Ti3+] = 5.0 x 10-2 M.
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 1.80 M and 0.130 M , respectively. The volume of half-cells is the same. a) What is the initial cell potential?    Express your answer using two significant figures. b) What is the cell potential when the concentration of Ni2+ has fallen to 0.600 M ?    Express your answer using two significant figures.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT