Respuesta :
Answer:
1. 0.97 V
2. [tex]Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)[/tex]
Explanation:
In this case, we can start with the half-reactions:
[tex]Ag^+~_(_a_q_)->~Ag_(_s_)[/tex]
[tex]Al_(_s_)~->~Al^+^3~_(_a_q_)[/tex]
With this in mind we can add the electrons:
[tex]Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)[/tex] Reduction
[tex]Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~[/tex] Oxidation
The reduction potential values for each half-reaction are:
[tex]Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)[/tex] - 0.69 V
[tex]Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)[/tex] -1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:
[tex]Al_(_s_)~->~Al^+^3~+~2e^-~[/tex] +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:
[tex]Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)[/tex]
I hope it helps!
