A galvanic cell is constructed using a hypothetical metal M and Silver ( Ag ) at 298 K . The cell is represented as: Pt | M ²⁺(0.01 M ), M ³⁺(0.1 M ) || Ag ⁺(0.01 M ) | Ag Given the standard reduction potentials: E^ _ Ag ⁺/ Ag = 0.80 V E^ _ M ²⁺/ M = -0.50 V E^ _ M ³⁺/ M = -0.10 V Taking 2.303RT F = 0.06 V , the EMF of the cell is:

Step-by-step solution

First, calculate the standard reduction potential for the M ³⁺/ M ²⁺ half-cell using G^ = -nFE^ . 1) M ³⁺ + 3 e ⁻ M G^ ₁ = -3F(-0.10) = 0.30F 2) M ²⁺ + 2 e ⁻ M G^ ₂ = -2F(-0.50) = 1.00F Subtracting (2) from (1) gives: M ³⁺ + e ⁻ M ²⁺ G^ ₃ = G^ ₁ - G^ ₂ = 0.30F - 1.00F = -0.70F Since G^ ₃ = -1FE^ _ M ³⁺/ M ²⁺ , we have: -1FE^ _ M ³⁺/ M ²⁺ = -0.70F E^ _ M ³⁺/ M ²⁺ = 0.70 V The overall cell reaction is: Anode: M ²⁺ M ³⁺ + e ⁻ Cathode: Ag ⁺ + e ⁻ Ag Overall: M ²⁺ + Ag ⁺ M ³⁺ + Ag The standard cell potential is: E^ _ ce