Chemistry - Electrochemistry Question with Solution | TestHub

Given, $\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-};E^{\circ }=+0.76\mathrm{V}$ $$\mathrm{Fe}⟶{\mathrm{Fe}}^{2+}+2e^{-};E^{\circ }=+0.41\mathrm{V}$$ On reversing the above equation (i) and (ii), we get ${\mathrm{Zn}}^{2+}+2e^{-}⟶\mathrm{Zn};E^{\circ }=-0.76\mathrm{V}$ ${\mathrm{Fe}}^{2+}+2e^{-}⟶\mathrm{Fe};E^{\circ }=-0.41\mathrm{V}$ [where, $E^{\circ }=$ standard reduction potential] To find, the standard emf of the cell with the reaction.So, $E_{\text{cell }}^{\circ }=E_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\circ }-E_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\circ }$ $=-0.41\mathrm{V}+0.76\mathrm{V}=+0.35\mathrm{V}$