Given the equilibrium constant :
`K_(c)` of the reaction : `Cu(s) + 2Ag^(+)(aq) rightarrow Cu^(2+) (Aq) + 2Ag(s)` is `10xx10^(15)`, calculate the `E_(cell)^(@)` of this reaction at 298K . `([2.303 (RT)/(Ft) at 298K = 0.059V])`

Options

A. `0.4736mV`
B. `0.04736mV`
C. `0.04736mV`
D. `0.4736V`

Explanation

To calculate the standard cell potential \( E_{cell}^\circ \) of the reaction given the equilibrium constant \( K_c \), we can use the Nernst equation. The reaction is:

\[ Cu(s) + 2Ag^+(aq) \rightleftharpoons Cu^{2+}(aq) + 2Ag(s) \]

### Step-by-Step Solution:

1. **Identify the Number of Electrons Transferred**:
The reaction shows that 2 moles of silver ions \( Ag^+ \) are reduced to 2 moles of silver \( Ag \), while copper \( Cu \) is oxidized from 0 to +2 oxidation state. Therefore, the total number of electrons transferred in the reaction is 2.
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