What is potential of platinum wire dipped into a solution of 0.1 M in Sn2+ and 0.01 M in Sn4+ ?

Question

What is potential of platinum wire dipped into a solution of 0.1 M in Sn2+ and 0.01 M in Sn4+ ? (A)  E°  (B)  E° +0.059  (C)  E° +(0.059/2)  (D)  E° -0.059

Tardigrade Admin · Accepted Answer

Sn2+ xrightarrowSn4++2e- Ecell=E° -(0.059/n) log ([Sn4+]/[Sn2+]) =E° -(0.059/2) log [ (0.01/0.1) ] =E° -(0.059/2)

Q. What is potential of platinum wire dipped into a solution of $0.1 M in S n^{2 +}$ and $0.01 M in S n^{4 +}$ ?

$E^{\circ}$

$E^{\circ} + 0.059$

$E^{\circ} + \frac{0.059}{2}$

$E^{\circ} - 0.059$

$E^{\circ} - 2 \times 0.59$

$S n^{2 +} S n^{4 +} + 2 e^{-}$ $E_{ce ll} = E^{\circ} - \frac{0.059}{n} lo g \frac{[ S n ^{4 +} ]}{[ S n ^{2 +} ]}$ $= E^{\circ} - \frac{0.059}{2} lo g [\frac{0.01}{0.1}]$ $= E^{\circ} - \frac{0.059}{2}$

Q. What is potential of platinum wire dipped into a solution of 0.1MinSn2+ and 0.01MinSn4+ ?

E∘

E∘+0.059

E∘+20.059​

E∘−0.059

E∘−2×0.59

Sn2+​Sn4++2e− Ecell​=E∘−n0.059​log[Sn2+][Sn4+]​ =E∘−20.059​log[0.10.01​] =E∘−20.059​