Chemistry - Electrochemistry Question with Solution | TestHub

$E_{cell}=E_{(Q,2H^{+}\mid H_{2}O)}-E_{SCE}$

Where,

$\begin{array}{rl}{E}_{(Q,2H^{+}/H_{2}O)}& =E_{(Q,2H^{+}/H_{2}O)}^o-0.059pH\\ \therefore E_{cell}& =\left[\left(E_{(Q,2H^{+}/H_{2}O)}^o-0.059pH\right)-E_{SCE}\right]\\ & =(0.7-0.059pH)-(0.24V)\\ & =(0.7-0.059\times 10)-(0.24V)=-0.13V\end{array}$

Because $E_{cell}$ is -ve, the reaction is endergonic (i.e., $\Delta G=+ve$).