Chemistry - Electrochemistry Question with Solution | TestHub

Cell equation : $A+2B^{+}⟶A^{2+}+2B$ $A^{2+}+2e^{-}⟶A;E^{\circ }=+0.34\mathrm{V}$ and ${\log }_{10}K=15.6$ at $300\mathrm{K}$ for cell reactions. Find $E^{\circ }$ for $B^{+}+e^{-}⟶B$. $[$ Given $:\frac{2.303RT}{nF}=0.059$ at $300\mathrm{K}]$