Question

(a) Calculate the mass of copper produced by the reduction of $Cu^{2+}$ ion at the cathode during the passage of 1.60 amperes of current through a solution of copper (ii) sulphate for 1 hour. $[Cu = 63.5, F = 96500C \text{mol}^{-1}]$ (b) Given that, $Ag^+ + e^- \rightarrow Ag(s)$; $E^o = +0.800V$ and $Mn^{2+} + 2e^- \rightarrow Mn(s)$; $E^o = -1.18 V$ Use the data to calculate the voltage, $E^o$ of the cell: $Mn(s)/Mn^{2+} (0.400M) // Ag^+(0.150M)/Ag(s)$

          (a)
Calculate the mass of copper produced by the reduction of $Cu^{2+}$ ion at the cathode
during the passage of 1.60 amperes of current through a solution of copper (ii)
sulphate for 1 hour.
$[Cu = 63.5, F = 96500C \text{mol}^{-1}]$
(b)
Given that, $Ag^+ + e^- \rightarrow Ag(s)$;     $E^o = +0.800V$ and
$Mn^{2+} + 2e^- \rightarrow Mn(s)$;     $E^o = -1.18 V$
Use the data to calculate the voltage, $E^o$ of the cell:
$Mn(s)/Mn^{2+} (0.400M) // Ag^+(0.150M)/Ag(s)$

(a)
Calculate the mass of copper produced by the reduction of Cu^2+ ion at the cathode
during the passage of 1.60 amperes of current through a solution of copper (ii)
sulphate for 1 hour.
[Cu = 63.5, F = 96500C mol^-1]
(b)
Given that, Ag^+ + e^- → Ag(s); E^o = +0.800V and
Mn^2+ + 2e^- → Mn(s); E^o = -1.18 V
Use the data to calculate the voltage, E^o of the cell:
Mn(s)/Mn^2+ (0.400M) // Ag^+(0.150M)/Ag(s)

Added by Paula T.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Mahendra Rathore

Step 1

Given: Current (I) = 160 amperes Time (t) = 1 hour = 3600 seconds Charge (Q) = I * t = 160 A * 3600 s = 576000 C Using Faraday's constant (F = 96500 C/mol), we can calculate the number of moles of electrons transferred: Moles of electrons (n) = Q / F = 576000 C Show more…

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Calculate the mass of copper produced by the reduction of Cu2+ on the cathode during the passage of 160 amperes of current through a solution of copper (II) sulfate for 1 hour. Co = 63.5 F = 96500 C/mol. Given that A + eA(s) = +0.800 V and M2+ + 2e = Mn(s) = -0.18 V, use the data to calculate the voltage of the cell Mn(s) | Mn(0.400 M) || Ag(s).