Question

1. Consider a voltaic cell in which the following reaction takes place in basic medium at 25℃.

2 NO3- (aq) + 3 S2- (aq) + 4 H2O 🡪 3 S (s) + 2 NO (g) + 8 OH- (aq)

1. Calculate E◦
2. Write the Nernst equation for the cell voltage E.
3. Calculate E under the following conditions: PNO = 0.994 atm, pH =13.7, [S2-] = 0.154 M, [NO3-] = 0.472 M

Answer 1

1. E⁰ = E$_1^⁰$ + E$_2^⁰$

E⁰ = 0.78 + 0.407

E⁰ = 1.187V

2. E = E⁰ - $\frac{0.059}{n}$log$\frac{[OH^-]⁸(p_{NO})^2}{[NO_3^{-2}]^2[S^{2-}]^³}$

E = E⁰ - $\frac{0.059}{6}$log$\frac{[OH^-]⁸(p_{NO})^2}{[NO_3^{-2}]^2[S^{2-}]^³}$

3. pH = 13.7

pOH = 0.3

[OH^-] = 10^-0.3 = 0.5M

[NO$_3^{-2}$ ] = 0.472M

[S^2-] =0.154M

p_NO = 0.994 atm

E = 1.187 - $\frac{0.059}{6}$log$\frac{(0.5)⁸(0.994)²}{(0.472)²(0.154)³}$

E = 1.187 + $\frac{0.059}{6}$ ×11 log 8.18

E = 1.187 + $\frac{0.059}{6}$ × 0.912

E = 1.187 + 0.098

E = 1.285 V