What is the standard electrode potential for the reaction 2 Cr + 3 Pb²⁺ → 3 Pb + 2 Cr³⁺?

Question

Pb^2+(aq) + 2e^- -> Pb(s) // -0.13

Cr^3+(aq) + 3e^- -> Cr(s) // -0.74

Answer 1

Pb would be the cathode (reduction)

Cr would be the anode (oxidation)

Eºcell = cathode - anode = -0.13 - (-0.74) = 0.61 V