The question is below it was just easier to take a screenshot.

In this situation we need to balance the two half cell equations to form a proper reaction. This means reversing one of the two half-cell equations and balancing.

In this case we do not need to actually worry about balancing because the electrical potentials for each half reaction remain unchanged regardless of reactant and product quantities. What does change however, is the sign of the electrical potential. As we are reversing one half-cell reaction, it's electrical potential will be reversed aswell. Then the overall cell potential is the sum of these two potentials.

The anode reaction is Fe3+ + e -> Fe2+ as it requires electrons as reactants and as a positive electrical potential, meaning this reaction favours the products. The cathode reaction is therefore manganese.

So the overall electrical potential Ecell = Ecathode - Eanode
Ecell = -1.18V - 0.77 = -1.95V

(remember we switch the polarity of the electrical potential for the reaction that we reverse, in this case the anode reaction involving iron).

The answer is d.potassium

Ametal with a large number of free flowing electrons most likely has good conductivity