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प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
Pt(s) | Br− (0.010 M) | Br2(l) || H+ (0.030 M) | H2(g) (1 bar) | Pt(s)
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उत्तर
The cell reaction is as follows:
\[\ce{2Br^- (0.010 M) + 2H^+ (0.030 M) -> Br2_{(l)} + H2 (1 bar)}\]
Hence, n = 2,
According to the Nernst equation for the cell, the emf is given below:
\[\ce{E_{cell} = (E^{\circ}_{H^+/\frac{1}{2}H_2} - E^{\circ}_{\frac{1}{2}Br_2/Br^-}) - \frac{0.0591}{2} log_10 \frac{pH_2}{[Br^-]^{2}[H^+]^2}}\]
= \[\ce{[0 - (+1.08)] - \frac{0.0591}{2} log_10 \frac{1}{(0.010)^2(0.030)^2}}\]
= \[\ce{-1.08 - \frac{0.0591}{2} log_10(1.11 \times 10^7)}\]
= −1.08 − 0.208
= −1.288 V
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