A voltaic cell utilizes the following reaction and operates at 298 K: 3Ce4+( a q

Question

A voltaic cell utilizes the following reaction and operates at 298 K:
3Ce4+(aq)+Cr(s)3Ce3+(aq)+Cr3+(aq).

Part A

What is the emf of this cell under standard conditions?

Express your answer using three significant figures.

Part B

What is the emf of this cell when [Ce4+]= 2.1 M , [Ce3+]= 0.18 M , and [Cr3+]= 1.1×102 M ?

Express your answer using two significant figures.

Part C

What is the emf of the cell when [Ce4+]= 1.7×102 M ,[Ce3+]= 2.7 M , and [Cr3+]= 1.6 M ?

Express your answer using two significant figures.

Explanation / Answer

The given reaction is

3 Ce4+ (aq) + Cr (s) ------> 3 Ce3+ (aq) + Cr3+ (aq)

The half reactions are:

Ce4+ + e- -------> Ce3+ ; E0 = +1.61 V (reduction half)

Cr3+ + 3e- ---------> Cr; E0 = -0.74 V (oxidation half)

The emf of the cell under standard conditions is given as

E0cell = E0red – E0ox = (+ 1.61 V) – (- 0.74 V) = 2.35 V

The equilibrium constant for the given reaction is given as

K = [Ce3+]3[Cr3+]/[Ce4+]3 (concentration of pure substances are neglected)

The emf of the cell under non-standard conditions is given as

Ecell = E0cell – 2.303.R.T/n.F(log10K) where T = absolute temperature; F = Faraday = 96500 Coulomb, n = number of moles of electrons transferred (3 in this case) and R is the gas constant.

Therefore,

Ecell = (2.35 V) – 2.303*(8.314 J/mol.K)*(298 K)/(3 mol)(96500 C) log10(0.18)3(1.1*10-2)/(2.1)3

or, Ecell = (2.35 V) – (0.0197 V).log10(5.832*10-3)(1.1*10-2)/9.261

or, Ecell = (2.35 V) – (0.0197 V).(-5.1594) = 2.35 V – (-0.1016 V) = 2.4516 V

The emf of the cell is 2.45 V (correct to 2 decimal places)/2.4 V (if you ask for 2 sig figures) (ans)

For the next part, we use the same equation as above and obtain

Ecell = E0cell – (0.0197 V).log10(2.7)3(1.6)/(1.7*10-2)3 =(2.35 V) – (0.0197 V)(6.8068)

= (2.35 V) – (0.1340 V) = 2.216 V

The emf of the cell is 2.22 V (correct to 2 decimal places)/ 2.2 V (2 sig figures) (ans)

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