Question

Use half-cell potentials to calculate `K_(eq)` for the reaction below?

`2Cu` (s)`+2H^+` (aq)`\toH_2` (g)

`E°_{Cu\toCu^(2+)}=-0.34` V

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My work:

Answer 1

`K_(eq) = 2.16xx10^-12`

Explanation:

`Cu^o(s) + 2H^+(aq) => Cu^(+2)(aq) + H_2(g)`

Using `E^o = (0.0592/n)logK_(eq)` => `log K_(eq) = ((n*E^o)/0.0592)`

For above reaction:
n = 2 and
`E^o(cell) = E^o("Reduction") - E^o("Oxidation")` = (0.00 volt) - (+0.34 volt) = -0.34 volt

`logK_(eq) = (2(-0.34))/0.0592 = -11.5`

`K_(eq) = 10^(-11.5) = 3.16xx10^-12`

Answer 2

Well, for this nonspontaneous reaction... `K = 3.16 xx 10^(-12)`.

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First of all, let's re-balance the reaction, which should be:

`"Cu"(s) -> "Cu"^(2+)(aq) + 2e^(-)`, `E_(red)^@ = +"0.34 V"`
`ul(2"H"^(+)(aq) + 2e^(-) -> "H"_2(g))`, `E_(red)^@ = "0.00 V"`
`2"Cu"(s) + 2"H"^(+)(aq) -> 2"Cu"^(2+)(aq) + "H"_2(g)`

The reduction potential of copper cation is more positive, so copper would rather be reduced.

Hence, this reaction is nonspontaneous as-written if we try our hardest and certainly fail to oxidize copper using `"HCl"`.

The (nonspontaneous) cell potential is:

`E_(cell)^@`

`= overbrace(E_(red)^@)^"reduction" +overbrace(E_(o x)^@)^"oxidation"`, OR `overbrace(E_("cathode")^@)^"reduction" - overbrace(E_"anode"^@)^"reduction"`

`= "0.00 V" + (-"0.34 V")`, OR `"0.00 V" - (+"0.34 V")`

`= -"0.34 V"`

From this we obtain the positive `DeltaG_(rxn)^@`:

`DeltaG^@ = -nFE_(cell)^@`

`= -(2 cancel("mol e"^(-)))/("1 mol Cu") cdot "96485 C/"cancel("mol e"^(-)) cdot (-"0.34 V")`

`=` `"65610 J/mol"`

As a result, since we are at equilibrium,

`cancel(DeltaG)^(0) = DeltaG^@ + RTlncancelQ^K`

and thus, we get the physically impractical equilibrium constant:

`color(blue)(K) = e^(-DeltaG^@//RT)`

`= e^(-(+"65610 J/mol")//("8.314 J/mol"cdot"K" cdot "298 K")`

`= e^(-26.5)`

`= color(blue)(3.16 xx 10^(-12))`