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Q.

A cell is containing two H electrodes. The negative electrode is in contact with a solution of pH = 6. EMF of the cell is 0.118 V at 25⁰C. Calculate pH at positive electrode

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answer is 4.

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Detailed Solution

$E_{cell} = 0 .059 \log \frac{[H^{+}] cathode}{[H^{+}] anode} = 0 .059, [pH anode - pH cathode] 0 .118 = 0 .059 [6 - pH] pH = 4$

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