Question

What is the voltage of the electrochemical cell written as: Cu(s) | Cu2+(aq) || Mg2+(aq) | Mg(s)?
A. 2.71 V
B. -2.71 V
C. -2.03 V
D. 2.03 V

Answer 1

Add up the reduction potentials:

\(\sf \xi_{cell}=\xi_{anode}+\xi_{cathode}\)

Answer 2

hold on i think i know

Answer 3

so it'd be like cu(s) = cu2+ +2e-

Answer 4

yes, Cu(s) is acting as the anode

Answer 5

+0.34-(-2.37)=0.34

Answer 6

2.71

Answer 7

so A right?

Answer 8

i dont know which value is which

Answer 9

sorry i'll retype it.

Answer 10

The potentials you're given are reduction potentials, you have to reverse one of them, then add them together.

Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V

Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V

Since Cu is acting as the anode, the equation needs to be reversed.

Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V

Ecell= -2.37 V+ (- 0.34 V) = -2.71 V

It makes sense that it's negative because Mg is more reactive than Cu, so it's not spontaneous.

Answer 11

oh okay. i see my mistake. you were correct. thank you for the explanation

Answer 12

no problem! i hope it's clearer

Answer 13

yea i understand now. big help thanks!