What must be the concentration of Ag^{+} in a | vedclass

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(C) For simultaneous deposition,the reduction potentials of both ions must be equal at the cathode.Using the Nernst equation for both half-reactions:$

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nearly $10^{-8} \ M$

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(C) For simultaneous deposition,the reduction potentials of both ions must be equal at the cathode.Using the Nernst equation for both half-reactions:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} + \frac{0.0591}{2} \log [Cu^{2+}]$$E_{Ag^{+}/Ag} = E^0_{Ag^{+}/Ag} + 0.0591 \log [Ag^{+}]$Equating the two potentials:$0.34 + \frac{0.0591}{2} \log(1.0) = 0.812 + 0.0591 \log [Ag^{+}]$Since $\log(1.0) = 0$,the equation becomes:$0.34 = 0.812 + 0.0591 \log [Ag^{+}]$$0.0591 \log [Ag^{+}] = 0.34 - 0.812 = -0.472$$\log [Ag^{+}] = \frac{-0.472}{0.0591} \approx -7.986 \approx -8$$[Ag^{+}] = 10^{-8} \ M$

What must be the concentration of $Ag^{+}$ in an aqueous solution containing $Cu^{2+} = 1.0 \ M$ so that both the metals can be deposited on the cathode simultaneously? Given that $E^0_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^0_{Ag^{+}/Ag} = 0.812 \ V$ at $T = 298 \ K$.

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