Question

Calculate the${\mathbf{emf}}_{\mathbf{x}}$of a dry cell for which a potentiometer is balanced when${\mathbf{R}}_{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{200}\mathbf{\Omega }$, while an alkaline standard cell with an emf $\mathbf{1}\mathbf{.}\mathbf{600}\mathbf{V}$of requiresto ${\mathbf{R}}_{\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{247}\mathbf{\Omega }$balance the potentiometer.

Short answer

The of a dry cell being for which a potentiometer is balanced is ${\mathrm{E}}_{\mathrm{x}}=1.540\mathrm{V}$.

Step-by-step solution

Concept Introduction

The electric potential generated by an electrochemical cell or a changing magnetic field is known as electromotive force. Electromotive force (EMF) is a well-known abbreviation.

Information Provided

• Emf of a standard cell:${\mathrm{E}}_{\mathrm{s}}=1.600\mathrm{V}$
• Resistance of the potentiometer:${\mathrm{R}}_{\mathrm{x}}=1.200\mathrm{\Omega }$
• Standard resistance value:${\mathrm{R}}_{\mathrm{s}}=1.247\mathrm{\Omega }$

Calculation for EMF of the cell

Calculate of a cell using a potentiometer. The formula is -

$$\begin{gathered} {\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{s}}\frac{{\mathrm{R}}_{\mathrm{x}}}{{\mathrm{R}}_{\mathrm{s}}} \\ =(1.600\mathrm{V})\frac{1.200\mathrm{\Omega }}{1.247\mathrm{\Omega }} \\ =1.600\mathrm{V}\times 0.96 \\ =1.540\mathrm{V} \end{gathered}$$

Therefore, the value for emf is ${\mathrm{E}}_{\mathrm{x}}=1.540\mathrm{V}$.