Question

The electrode oxidation potential of electrode $M\left(s\right)\to M^{n+}(aq.)\left(2M\right)+n{e}^{-}$ at 298K is $E_1$. When temperature (in $C^{\circ }$) is doubled and concentration is made half, then the electrode potential becomes $E_2$. Which of the following represent the correct relationship between $E_1$ and $E_2$?

• A. $E_1>E_2$
• B. $E_1<E_2$
• C. $E_1=E_2$
• D. Can’t be predicted

Answer 1

The correct option is B $E_1<E_2$

Use Nernst equation,

$E_{cell}=E_{cell}^0-\frac{2.303RT}{nF}log\left[M^{n+}\right]$

$E_1=E^0-\frac{\mathrm{2.303.}RT}{nF}log\left[2\right]$

$E_2=E^0-\frac{\mathrm{2.303.}R\left(2T\right)}{nF}log\left[1\right]$

$E_2=E^0$

$Sincelog\left(1\right)=0$

$E_1-E_2=-\frac{\mathrm{2.303.}RT}{nF}log\left[2\right]$

$E_1=E_2-\frac{\mathrm{2.303.}RT}{nF}log\left[2\right]$

Therefore $E_2>E_1$