Given below are the half-cell reactions: Mn ²⁺ + 2e^- Mn (E^0 = -1.18 V ) 2( Mn ³⁺ + e^- Mn ²⁺) (E^0 = +1.51 V ) The E^0_ cell for 3 Mn ²⁺ Mn + 2 Mn ³⁺ will be _____________.

Step-by-step solution

The given target reaction is 3 Mn ²⁺ Mn + 2 Mn ³⁺ This reaction can be split into two half-reactions: Cathode (Reduction): Mn ²⁺ + 2e^- Mn Anode (Oxidation): 2 Mn ²⁺ 2 Mn ³⁺ + 2e^- The standard reduction potentials are: E^0_ cathode = E^0_ Mn ²⁺/ Mn = -1.18 V E^0_ anode = E^0_ Mn ³⁺/ Mn ²⁺ = +1.51 V The standard cell potential is given by: E^0_ cell = E^0_ cathode - E^0_ anode E^0_ cell = -1.18 V - 1.51 V = -2.69 V Since E^0_ cell is negative, the standard Gibbs free energy change G^0 = -nFE^0_ cell is positive. Th