Three metals P , Q , and R form two different galvanic cells under standard conditions. The standard Gibbs free energy changes ( G ^ ) for the cells are given as: Cell 1: P | P ²⁺(1 M ) | | Q ²⁺(1 M ) | Q , G ^ =-193 ~kJ ~mol ⁻¹ Cell 2: Q | Q ²⁺(1 M ) | | R ³⁺(1 M ) | R , G ^ =-289.5 ~kJ ~mol ⁻¹ If the standard reduction potential of P ²⁺ / P is -0.40 ~V , what is the standard reduction potential of R ³⁺ / R ? (Given

Step-by-step solution

For Cell 1, the cell reaction is P + Q ²⁺ P ²⁺ + Q . The number of electrons transferred is n=2 . Using G ^ = - nFE _ cell ^ : -193000 = -2 96500 E _ cell1 ^ E _ cell1 ^ = 1.00 ~V Since E _ cell1 ^ = E _ Q ²⁺/ Q ^ - E _ P ²⁺/ P ^ : 1.00 = E _ Q ²⁺/ Q ^ - (-0.40) E _ Q ²⁺/ Q ^ = 0.60 ~V For Cell 2, the cell reaction is 3 Q + 2 R ³⁺ 3 Q ²⁺ + 2 R . The total number of electrons transferred is n=6 . Using G ^ = - nFE _ cell ^ : -289500 = -6 96500 E _ cell2 ^ E _ cell2 ^ = 289500 579000 = 0.50 ~V Since E _ cell2 ^ = E _