Question

Aluminium displaces hydrogen from acids,but copper does not. A galvanic cell prepared by combining Cu/Cu2+ and Al/Al3+ has an EMF of 2.0 V at 298 K. If the potential of copper electrode is + 0.34 V,that of aluminium electrode is

a)-2.3 V

b)+ 2.34 V

c)- 1.66 V

d)1.66 V

Correct answer is option 'C'. Can you explain this answer?​

Answer 1

Answer:

Explanation:

That was the correct answer

I hope you will understand

Answer 2

Answer:

The potential of Aluminum electrode is -1.66V i.e.option(c).

Explanation:

The EMF of the cell is given as,

$E=E_c-E_a$         (1)

Where,

E=emf of the cell

Ec=electrode potential of the cathode

Ea=electrode potential of the anode

As aluminum displaces hydrogen from acids but copper does not which indicates that aluminum is more reducing than copper. That means that aluminum will act as a reducing agent and its oxidation will take place at the anode while the reduction of copper will take place at the cathode.

From the question we have,

$E_c=0.34V$ and $E=2V$

By substituting the values E and Ec in equation (1) we get;

$2=0.34-E_a$

$E_a=0.34-2=-1.66V$

Hence,the potential of Aluminum electrode is -1.66V i.e.option(c).