An electrochemical cell is constructed using electrodes with the following half-cell reactions.

E cell = +1.95 V

Explanation:

At Anode : Oxidation reaction takes place

At Cathode : Reduction reaction takes place

The reaction with lower value of reduction potential will undergo Oxidation

  E = -1.18 V

This equation undergo oxidation reaction and become:

Anode(Oxidation-Half) :

     E = +1.18 V

Cathode(Reduction-Half) :

   E =+0.77 V

To balance the reaction multiply reduction-Half with 2.We get :

Note that E is intensive property , do not multiply E  of oxidation-half with 2

Ecell =  0.77 -(-1.18)

E = +1.95 V

the answer would be a

i believe the answer is true.