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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 .

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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

Ecell = 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(Ecell) will increase

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