What is standard reduction potential of Cu2+ | Cu(s) if E° of following cell is 0.46V ? Cu(s) | Cu2+(aqs) || Ag+(aq) | Ag(s) ( Eo = 0.80V)
The overall cell potential, E°cell, is the difference between the reduction and oxidation potentials:
E°cell = E°reduction - E°oxidation
Given E°cell = 0.46V, and E°reduction = 0.80V,
we can rearrange the equation to solve for E°oxidation:
E°oxidation = E°reduction - E°cell
E°oxidation = 0.80V - 0.46V = 0.34V
Therefore, the standard reduction potential of Cu²⁺ | Cu(s) is 0.34V. The correct option is: (D) 0.34 V


Electricity is passed through an acidic solution of Cu$^{2+}$ till all the Cu$^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ___ mL. (Nearest integer)
Given:
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)}$
$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$
Faraday constant = 96500 C mol$^{-1}$
Molar volume at STP = 22.4 L
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. It converts chemical energy to electrical energy.
It consists of two half cells and in each half cell, a suitable electrode is immersed. The two half cells are connected through a salt bridge. The need for the salt bridge is to keep the oxidation and reduction processes running simultaneously. Without it, the electrons liberated at the anode would get attracted to the cathode thereby stopping the reaction on the whole.