# Question #a3de5

May 27, 2017

The arrangements in Daniel cell

Anode reaction

$Z n \left(s\right) \to Z {n}^{2 +} \left(a q\right) + 2 {e}^{-}$

Cathode reaction

$C {u}^{2 +} \left(a q\right) + 2 {e}^{-} \to C u \left(s\right)$

Overall reaction

$Z n \left(s\right) + C {u}^{2 +} \left(a q\right) \to C u \left(s\right) + Z {n}^{2 +} \left(a q\right)$

The cell notation

$Z n \left(s\right) | Z {n}^{2 +} | | C {u}^{2 +} | C u \left(s\right)$

The Cell Emf by Nernst equation

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \times \frac{\left[Z {n}^{2 +}\right]}{\left[C {u}^{2 +}\right]} \ldots \ldots \left[1\right]$

If we add sodium sulphide $\left(N {a}_{2} S\right)$ to cupper sulphate in daniel cell, it interacts with $C {u}^{2 +}$ in solution and lowers the concentration of $C {u}^{2 +}$ by precipitating it as insoluble $C u S \left(s\right)$ as per the following equation.

$C {u}^{2 +} \left(a q\right) + N {a}_{2} S \to C u S \left(s\right) \downarrow + 2 N {a}^{+}$

As a result the ratio $\frac{\left[Z {n}^{2 +}\right]}{\left[C {u}^{2 +}\right]}$ in equation [1] increases and this decreases the ${E}_{c e l l}$ to a great extent.