The standard reduction potentials for two half-cells are given as: E^ _ Cr ³⁺/ Cr = -0.74 V E^ _ Cd ²⁺/ Cd = -0.40 V What is the standard Gibbs free energy change ( G^ ) for the spontaneous cell reaction formed by these two half-cells? (Given: 1 F = 96500 C mol ⁻¹ )

Step-by-step solution

For a spontaneous cell reaction, the standard cell potential ( E^ _ cell ) must be positive. The half-cell with the higher standard reduction potential undergoes reduction (cathode), and the one with the lower standard reduction potential undergoes oxidation (anode). Here, E^ _ Cd ²⁺/ Cd (-0.40 V ) > E^ _ Cr ³⁺/ Cr (-0.74 V ) . Thus, Cd ²⁺ is reduced and Cr is oxidized. E^ _ cell = E^ _ cathode - E^ _ anode = -0.40 V - (-0.74 V ) = +0.34 V The balanced cell reaction is: 2 Cr + 3 Cd ²⁺ 2 Cr ³⁺ + 3 Cd The number of e