Calculate the EMF of the Galvanic cell: $ \text{Zn} | \text{Zn}^{2+}(1.0 M) \parallel \text{Cu}^{2+}(0.5 M) | \text{Cu} $ Given: $ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.763 \, \text{V} $ and $ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.350 \, \text{V} $
\(Pt(s) ∣ H2(g)(1atm) ∣ H+(aq, [H+]=1)\, ∥\, Fe3+(aq), Fe2+(aq) ∣ Pt(s)\)
Given\( E^∘_{Fe^{3+}Fe^{2+}}\)\(=0.771V\) and \(E^∘_{H^{+1/2}H_2}=0\,V,T=298K\)
If the potential of the cell is 0.712V, the ratio of concentration of \(Fe2+\) to \(Fe3+\) is
A convex lens has power \( P \). It is cut into two halves along its principal axis. Further, one piece (out of two halves) is cut into two halves perpendicular to the principal axis as shown in the figure. Choose the incorrect option for the reported lens pieces.
The equation \[ 2 \cos^{-1} x = \sin^{-1} \left( 2 \sqrt{1 - x^2} \right) \] is valid for all values of \(x\) satisfying:
A metallic sphere of radius \( R \) carrying a charge \( q \) is kept at a certain distance from another metallic sphere of radius \( R_4 \) carrying a charge \( Q \). What is the electric flux at any point inside the metallic sphere of radius \( R \) due to the sphere of radius \( R_4 \)?
The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is:
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.