Question

Write the Nernst equation and emf of the following cell at 298 K:

Pt_(s) | Br⁻ (0.010 M) | Br_2(l) || H⁺ (0.030 M) | H_2(g) (1 bar) | Pt_(s)

Answer 1

The electrode reactions are:

At anode: \[\ce{2Br- (0.01 M) -> Br2_{(l)} + 2e-}\]

At cathode: \[\ce{2H+ (0.03 M) + 2e- -> H2 (1 bar)}\]

Net reaction: \[\ce{2Br^- (0.010 M) + 2H^+ (0.030 M) -> Br2_{(l)} + H2 (1 bar)}\]

Hence, n = 2,

According to the Nernst equation for the cell, the emf is given below:

E_cell =\[\ce{(E^{\circ}_{H^+/\frac{1}{2}H_2} - E^{\circ}_{\frac{1}{2}Br_2/Br^-}) - \frac{0.0591}{2} log_10 \frac{P_{H_2}}{[Br^-]^{2}[H^+]^2}}\]

= \[\ce{(0 - (+1.08)) - \frac{0.0591}{2} log_10 \frac{1}{(0.010)^2(0.030)^2}}\]

= `-1.08 - 0.0296 log_10 [1/((1 xx 10^-4) (9 xx 10^-4))]`

= `-1.08 - 0.0296 xx log_10 (10^8/9)`

= −1.08 − 0.0296 (log₁₀ 10⁸ − log₁₀ 9) V

= −1.08 − 0.0296 (8 − 0.9542) V

= −1.08 − 0.0296 (7.0458)

= −1.08 − 0.21

= −1.288 V