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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) - Chemistry

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Question

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)

Numerical
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Solution

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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Chapter 2: Electrochemistry - Exercises [Page 59]

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NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 2 Electrochemistry
Exercises | Q 2.5 (iv) | Page 59
Nootan Chemistry Part 1 and 2 [English] Class 12 ISC
Chapter 3 Electrochemistry
'NCERT TEXT-BOOK' Exercises | Q 3.5 (iv) | Page 210

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