Question

QUESTION 15 Which of the following changes will increase the voltage of the cell Al(s)A3+(aq, 1.00 M )lCu2(aq, 1.00 M)|Cu(s)? I. Increasing the surface area of the Al electrode Il. Decreasing the concentration of A3+(aa) to 0.001 M lI. Decreasing the M concentration of Cu2 la) to 0.001 a. b. Ill only OG. II only o d. ll andI e1 and ll only

PLEASE EXPLAIN WHY .

Answer 1

Answer

c) II only

Explanation

Oxidation half reaction

2Al(s) --------> Al³⁺(aq) + 6e^-

Reduction half reaction

3Cu²⁺(aq) + 6e^- ---------> 3Cu(s)  

Overall reaction

2Al(s) + 3Cu²⁺(aq) -------> 2Al³⁺(aq) + 3Cu(s)

reaction quotient , Q = [Al³⁺]²/[Cu²⁺]³

number of electron transfer , n = 6

Nernst equation at 25℃ is

E_cell = E°_cell - (0.0592V/n)logQ

E°_cell and n is constant , the variable is Q

so , the voltage of the cell depends on reaction quotient Q

Q is depends on concentration of Al³⁺ and Cu²⁺

when Al³⁺ concentration decrease Q value decreases

As Q value decreases cell voltage(E_cell) will increase