Question

What is the standard cell potential for the reaction of silver cation with aluminum metal?

`A)` `"2.47 V"`
`B)` `"0.87 V"`
`C)` `"0.73 V"`
`D)` `"1.67 V"`

Answer 1

I get A.

---

As a short review, `E_"cell"` is the drop in voltage across two electrodes for a full electrochemical reaction, which is the sum of two half-reactions.

`E_"cell"^@` is the same thing, except it is at `25^@ "C"` and standard pressure (generally, `"1 atm"`), and `"1 M"` concentrations.

The most reasonable reaction to occur is the one in which `E_"cell"^@` is composed of elements more capable of being oxidized that are being oxidized, or elements more capable of being reduced that are being reduced.

From the activity series below:

we should notice that elements higher up are more easily oxidized. In order for the overall reaction to be spontaneous, it must be that aluminum is oxidized, since it is more easily oxidized.

Or, you may have been given a series of reduction half-reactions instead, in which case, if `bb(E_"red"^@)` is more positive, it means the substance is more easily reduced.

Therefore, we should flip the second reaction and get:

`3(Ag^(+)(aq) + e^(-) -> Ag(s))`, `E_"red."^@ = +"0.80 V"`

`-(Al^(3+)(aq) + 3e^(-) -> Al(s))`, `-E_"red."^@ = -(-"1.67 V")`

`"-------------------------------------------------"`

`3Ag^(+)(aq) + Al(s) -> 3Ag(s) + Al^(3+)(aq)`, `E_"cell"^@ = ???`

Although `E` does change sign when one reverses a reaction, it is not an extensive property, and does not scale when one multiplies a reaction by a scalar.

So, `-E_"red."^@ = E_"ox."^@` for the second half-reaction, but we do not multiply `E_"red."^@` for the first reaction by `3`. That can be proven when we examine the answer choices.

If we did NOT scale the first reduction half-reaction by `3`:

`color(blue)(E_"cell"^@) = E_"cathode"^@ + E_"anode"^@`

`= E_"red."^@ + E_"ox."^@`

`= E_("red.","Ag")^@ - E_("red.","Al")^@`

`= "0.80 V" - (-"1.67 V") = color(blue)("2.47 V")`

which is A.

If we DID scale the first reduction half-reaction by `3`:

`cancel(color(red)(E_"cell"^@) = E_"red."^@ + E_"ox."^@)`

`= cancel(color(red)(3)E_("red.","Ag")^@ - E_("red.","Al")^@)`

`= cancel(3("0.80 V") - (-"1.67 V") = color(red)("4.07 V"))`,

which is not one of the answer choices.