Question

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

1) What would the overall potential for this cell be?

2) Write the standard cell notation for an electrochemical cell consisting of an anode and cathode of the same types as in this experiment, connected through a salt bridge.

Answer 1

1. 0.97 V

2.
`Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)`

Step-by-step explanation:

In this case, we can start with the half-reactions:

`Ag^+~_(_a_q_)->~Ag_(_s_)`

`Al_(_s_)~->~Al^+^3~_(_a_q_)`

With this in mind we can add the electrons:

`Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)` Reduction

`Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~` Oxidation

The reduction potential values for each half-reaction are:

`Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)` - 0.69 V

`Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)` -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

`Al_(_s_)~->~Al^+^3~+~2e^-~` +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

`Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)`

I hope it helps!