Question

Enter your answer in the provided box. ____ V |

Answer #1

For concentration cell, cathode and anode are same electrode

So, Eo = 0

Number of electron being transferred in balanced reaction is 2

So, n = 2

If E is positive anode will be the one with lower concentration

use:

E = Eo - (2.303*RT/nF) log {[Mg2+] at anode/[Mg2+]at cathode}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Mg2+] at anode/[Mg2+]at cathode}

E = 0 - (0.0591/2) log (0.33/0.54)

E = 6.323*10^-3 V

Answer: 6.32*10^-3 V

Calculate the emf of the following concentration cell: Mg(s) |
Mg2+(0.32 M) || Mg2+(0.45 M) | Mg(s) V

Enter your answer in the provided box.
What is the emf of a cell consisting of a Pb2+ / Pb
half-cell and a Pt / H+ / H2 half-cell if
[Pb2+] = 0.78 M, [H+] = 0.079
M and PH2 = 1.0 atm ?

For the following concentration cell: Mg(s) |
Mg2+(0.24 M) | Mg2+(0.53 M) | Mg(s)
(a) Determine the initial emf (Δε) for this cell (in volts)
(b) Determine the concentration of Mg2+ in each cell
when the emf of this cell goes to 0 volts
T = 298 K T = 310 K

Enter your answer in the provided box.
Calculate the following quantity: molarity of a solution
prepared by diluting 45.77 mL of 0.0371 M
ammonium sulfate to 450.00 mL.

Enter your answer in the provided box. Calculate the pH at the
equivalence point for the following titration: 0.40 M HCl versus
0.40 M methylamine (CH3NH2). The Ka of methylammonium is 2.3 ×
10−11

Enter your answer in the provided box.
The concentration of lead ions (Pb2+) in a sample of
polluted water that also contains nitrate ions
(NO3−) is determined by adding solid sodium
sulfate (Na2SO4) to exactly 500 mL of the
water. Calculate the molar concentration of Pb2+ if
0.00295 g of Na2SO4 was needed for the
complete precipitation of lead ions as PbSO4.
__________M
please help!!!!!

A concentration cell based on the following half reaction at 274
K
Mg2+ + 2 e- → Mg
SRP = -2.37 V
has initial concentrations of 1.31 M Mg2+, 0.239 M
Mg2+, and a potential of 0.02008 V at these conditions.
After 1.7 hours, the new potential of the cell is found to be
0.01032 V. What is the concentration of Mg2+ at the
cathode at this new potential?

A concentration cell has both electrodes made of magnesium and
is based on the following half-cell reaction.
Mg2+(aq) + 2 e− → Mg(s)
(a) If [Mg2+ ] = 2.18 M at the cathode and [Mg2+ ] = 0.00744 M
at the anode, what is the cell potential of the concentration cell
at 298.15 K?
(b) What is the cell potential of the cell when the
concentration of Mg2+ ions is the same in the cathode
and anode compartments?

When the Cu2+ concentration is
1.40 M, the observed cell potential at 298K for an
electrochemical cell with the following reaction is
2.805V. What is the
Mg2+ concentration?
Cu2+(aq) +
Mg(s) ----->
Cu(s) +
Mg2+(aq)
Answer: ______M

What is the calculated value of the cell potential at 298K for
an electrochemical cell with the following reaction, when the Cu2+
concentration is 1.44 M and the Mg2+ concentration is 1.47×10-4 M
?
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)
Answer: V
The cell reaction as written above is spontaneous for the
concentrations given:
_______
true
false

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