n identical cells, each of emf $ E $ and internal resistance $ r $, are joined in series to form a closed circuit. One of the cells, A, is joined with reversed polarity, the potential difference across each cell except A, is
(A) $ \dfrac{{2E}}{n} $
(B) $ \dfrac{{n - 1}}{n}E $
(C) $ \dfrac{{n - 2}}{n}E $
(D) $ \dfrac{{2n}}{{n - 2}}E $

Hint: The emf will be reduced due to the internal resistance of the identical cells. When one of the cells is connected with reversed polarity, it cancels out the effect of one other cell.

Formula used: In this solution we will be using the following formula;
 $ \sum V = 0 $ where $ V $ is the voltage dropped across the individual element.
 $ V = IR $ where $ V $ is the potential difference dropped across a resistor, and $ R $ is the resistance of the resistor, and $ I $ is the current flowing through the resistor.
 $ V = E - Ir $ where $ V $ is the potential difference across a cell, $ E $ is the emf of the cell, and $ r $ is the internal resistance of the cell.

Complete step by step answer
In the arrangement of the cell, the sum of the emfs gives the total emf of the circuit which would have been $ nE $ where $ E $ is the emf value and $ n $ is the number of cells. Since one is reversed, it would be negative. This negative value will cancel out another emf. Hence, the total becomes, $ nE - 2E $.
The potential difference across one cell is
 $ V = E - Ir $ where $ I $ is the current through the circuit, and $ r $ is the internal resistance.
By Kirchhoff’s voltage law, we can write
 $ E - Ir + E - Ir +... - E - Ir = 0 $
 $ \Rightarrow nE - 2E = Inr $
Hence, the current is given by
 $ I = \dfrac{{nE - 2E}}{{nr}} = \dfrac{{E(n - 2)}}{{nr}} $
Hence, by inserting the expression for the current $ V = E - Ir $ into the potential difference can be given as
 $ V = E - \dfrac{{E(n - 2)}}{{nr}}r = E\left[ {1 - \dfrac{{(n - 2)}}{n}} \right] $
By evaluating the bracket, we get
 $ V = E\left( {\dfrac{2}{n}} \right) = \dfrac{{2E}}{n} $

Hence, the correct option is A.

Note
For clarity, we could perform Kirchhoff's voltage law as $ E - Ir + E - Ir +... - E - Ir = 0 $ because the internal resistance $ r $ of a cell can be modelled to be a resistance connected in series to a perfect cell of emf $ E $.