Question

The equilibrium constant for the reaction
\(Zn(s)+Sn^{2+}(aq)⇌Zn^{2+}(aq)+Sn(s)\, is\, 1×10^{20}\) at \(298 K\). 
The magnitude of standard electrode potential of \(Sn/Sn^{2+}\) if \(E^{\degree}_{ Zn^{2+}/Zn}= –0.76V \) is _______ \(×10^{−2} V\). (Nearest integer).
(Given : \(\frac{2.303RT}{ F}\) \(=0.059 V \))

Answer 1

Correct Answer: 17

To find the standard electrode potential of \(Sn/Sn^{2+}\), we use the Nernst equation and the equilibrium constant: \(K\).
Given,\(K = 1 \times 10^{20}\), \(E^{\degree}_{ Zn^{2+}/Zn} = -0.76 \, V\), and \(\frac{2.303RT}{F} = 0.059 \, V\) at \(298K\).
For the reaction \(Zn(s) + Sn^{2+}(aq) \⇌ Zn^{2+}(aq) + Sn(s)\), the cell potential \(E^{\degree}_{cell}\) is given by:
\(E^{\degree}_{cell} = E^{\degree}_{Sn/Sn^{2+}} - E^{\degree}_{Zn^{2+}/Zn}\).
The Nernst equation relates the cell potential to the equilibrium constant:
\(E^{\degree}_{cell} = \frac{0.059}{2} \log K\) where the number of electrons transferred is 2.
Substitute values:
\(E^{\degree}_{cell} = \frac{0.059}{2} \log(1 \times 10^{20})\)
Calculate \(\log 10^{20} = 20\).
Thus, \(E^{\degree}_{cell} = \frac{0.059}{2} \cdot 20 = 0.59\, V\).
Using \(E^{\degree}_{cell} = E^{\degree}_{Sn/Sn^{2+}} - (-0.76)\):

\(0.59 = E^{\degree}_{Sn/Sn^{2+}} + 0.76\)
Solve for \(E^{\degree}_{Sn/Sn^{2+}}:\)

\(E^{\degree}_{Sn/Sn^{2+}} = 0.59 - 0.76 = -0.17\, V\)

Express as \(-17 \times 10^{-2}\, V\). The nearest integer for the magnitude is 17, within range (17,17).