Question

What will be the electrode potential for the given half cell reaction at pH= 5?

(R=8.314 J mol^-1K^-1; temp.=298 K; oxygen under std. atm. The pressure of 1 bar.)

Answer 1

Electrode potential

The electromotive force formed here between the standard reference electrode and another electrode to be identified is referred to as electrode potential in electrochemistry. The standard hydrogen electrode is used as the reference electrode by convention (SHE). It does, in fact, have a zero-volt potential.

Electrode potential explained in a picture

Given reaction at pH= 5, the solution is

$4\mathrm{H}++4\mathrm{e}-\to 2{\mathrm{H}}_2;{{\mathrm{E}}^0}_{\mathrm{red}}=0$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}^{\mathrm{o}}-\mathrm{n}\times 0.0591\log [\mathrm{ReactantProduct}]$

=$1.23-4\times 0.0591\log {[\mathrm{H}+]}^4{[{{\mathrm{P}}_{\mathrm{H}}}_2]}^2;[{{\mathrm{P}}_{\mathrm{H}}}_2=1\mathrm{bargiven}]$

$=1.{23}^{-4}\times 0.0591\times 4\times [-\log ({\mathrm{H}}^{+}\left)\right]$

But $\mathrm{pH}=5,{\mathrm{H}}^{+}={10}^{-5}\mathrm{M}$

∴${\mathrm{E}}_{\mathrm{cell}}=1.23-0.0591\times [-\log ({10}^{-5}\left)\right]$

$=1.23-0.0591\times 5\times 1$

$=1.23-0.2955$

$=0.9345\approx 0.93$

${\mathrm{E}}_{\mathrm{cell}}=0.93\mathrm{V}$