What will be the change in the electrode potential of chromium electrode dipping into chromic sulphate solution, when the electrolyte is diluted 10 times at 25^ C? [ E^0 ( Cr ³⁺/ Cr )] = -0.74 V

Step-by-step solution

The reduction half-reaction for the chromium electrode is: Cr ³⁺ + 3e^- Cr The Nernst equation for the reduction potential is given by: E = E^0 - 0.0591 n 1 [ Cr ³⁺] E = E^0 + 0.0591 3 [ Cr ³⁺] Let the initial concentration of Cr ³⁺ be c . The initial electrode potential is: E₁ = E^0 + 0.0591 3 c When the solution is diluted 10 times, the new concentration becomes c 10 . The new electrode potential is: E₂ = E^0 + 0.0591 3 ( c 10 ) The change in electrode potential is: E = E₂ - E₁ = 0.0591 3 [ ( c 10 ) - c ] E = 0.0