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Electrochemistry question

2024 · 27 Jan · Shift 2 · Q25
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Electrochemistry question

2024 · 27 Jan · Shift 2 · Q25

JEE MainChemistryElectrochemistryNumerical+4 / −1
The hydrogen electrode is dipped in a solution of pH=3\mathrm{pH}=3pH=3 at 25∘C25^{\circ} \mathrm{C}25∘C. The potential of the electrode will be ‾×10−2 V\underline{\hspace{2cm}}\times 10^{-2} \mathrm{~V}​×10−2 V. (2.303RTF=0.059 V)\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)(F2.303RT​=0.059 V)
Numerical answer
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