At 298 K a cell is built from two…

At 298 K a cell is built from two half-cells. Left (anode): a silver wire coated with solid AgBr dipping in 0.10 M KBr. Right (cathode): copper metal…

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298\ \mathrm{K}

a cell is built from two half-cells. Left (anode): a silver wire coated with solid

\mathrm{AgBr}

dipping in

0.10\ \mathrm{M}

\mathrm{KBr}

. Right (cathode): copper metal in

0.010\ \mathrm{M}

\mathrm{CuSO_4}

. Given

E^\circ(\mathrm{Ag^+|Ag})=0.80\ \mathrm{V}

E^\circ(\mathrm{Cu^{2+}|Cu})=0.34\ \mathrm{V}

K_{sp}(\mathrm{AgBr})=5.0\times10^{-13}

and

\frac{2.303RT}{F}=0.0591\ \mathrm{V}

, the emf of the cell is closest to

A. $0.15\ \mathrm{V}$
B. $0.21\ \mathrm{V}$
C. $0.27\ \mathrm{V}$
D. $0.35\ \mathrm{V}$

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Answer: (A) $0.15\ \mathrm{V}$

The

\mathrm{AgBr|Ag}

electrode is a silver electrode whose

[\mathrm{Ag^+}]=K_{sp}/[\mathrm{Br^-}]=5.0\times10^{-12}\ \mathrm{M}

. Hence

E=0.80+0.0591\log(5.0\times10^{-12})=0.80-0.6674=0.1326\ \mathrm{V}

(equivalently

E^\circ_{\mathrm{AgBr}}=0.073\ \mathrm{V}

and

E=0.073-0.0591\log0.10=0.132\ \mathrm{V}

). The copper electrode gives

E=0.34+\frac{0.0591}{2}\log(0.010)=0.34-0.0591=0.281\ \mathrm{V}

. Emf

=E_{cathode}-E_{anode}=0.281-0.132=0.149\approx0.15\ \mathrm{V}

. Using the

E^\circ

\mathrm{AgBr}

without the bromide term gives

0.21\ \mathrm{V}

, and adding instead of subtracting the

\log[\mathrm{Br^-}]

term gives

0.27\ \mathrm{V}

. Omitting the Nernst correction of the copper electrode gives

0.34-0.132=0.21

as well, so both electrodes need the full Nernst treatment.

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Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake