# Question & Answer: A concentration cell based on the following half reaction at 305 K Mg2+ + 2 e- → Mg SRP =…..

A concentration cell based on the following half reaction at 305 K Mg2+ + 2 e- → Mg SRP = -2.37 V has initial concentrations of 1.27 M Mg2+, 0.255 M Mg2+, and a potential of 0.02110 V at these conditions. After 10 hours, the new potential of the cell is found to be 0.009984 V. What is the concentration of Mg2+ at the cathode at this new potential?

$E_{cell} = \dfrac {0.0591}{n}log \dfrac {[Mg^{2+}] _{cathode}}{[Mg^{2+}] _{anode}}$

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$0.02110 V = \dfrac {0.0591}{2} log \dfrac { 1.27 }{0.255}$

$0.02110 \: V = 0.02110 \: V$

Thus electrode with $[Mg^{2+}] = 1.27 \: M$ is cathode and the electrode with $[Mg^{2+}] = 0.255 \: M$ is anode.

At cathode $[Mg^{2+}]$ will decrease to $\left ( 1.27-x \right ) \: M$ . At anode $[Mg^{2+}]$ will increase to $\left ( 0.255 + x \right ) \: M$ .

After 10 hours, the new potential of the cell is found to be 0.009984 V.

$E_{cell} = \dfrac {0.0591}{n}log \dfrac {[Mg^{2+}] _{cathode}}{[Mg^{2+}] _{anode}}$

$0.009984 V = \dfrac {0.0591}{2} log \dfrac { 1.27-x }{0.255+x}$

$log \dfrac { 1.27-x }{0.255+x} = 0.3379$

$\dfrac { 1.27-x }{0.255+x} =2.177$

$1.27-x =2.177 \left ( 0.255+x \right )$

$1.27-x = 0.555+2.177x$

$1.27 = 0.555+3.177x$

$3.177x=0.7149$

$x=0.225$

$[Mg^{2+}] = 1.27-x = 1.27 - 0.225 = 1.05 \: M$

This is the concentration of $Mg^{2+}$ at the cathode at this new potential