Solution:

We are tasked to find the external resistance $R$ when the potential difference across the first cell is zero. Let the emf of each cell be $E$, the internal resistances of the two cells be $r_1$ and $r_2$, and the external resistance be $R$.

Key points:

1. Current in the circuit: The total resistance in the circuit is $r_1 + r_2 + R$. The current $I$ is given by:

   $$I = \frac{E + E}{r_1 + r_2 + R} = \frac{2E}{r_1 + r_2 + R}.$$

2. Potential difference across the first cell: The potential difference across the first cell is:

   $$V_1 = E - I r_1.$$Since the potential difference across the first cell is zero, we set $V_1 = 0$:

   $$0 = E - I r_1.$$Substitute $I = \frac{2E}{r_1 + r_2 + R}$:

   $$0 = E - \frac{2E}{r_1 + r_2 + R} \cdot r_1.$$

3. Simplify the equation: Rearrange:

   $$E = \frac{2E r_1}{r_1 + r_2 + R}.$$Divide through by $E$ (since $E \neq 0$):

   $$1 = \frac{2r_1}{r_1 + r_2 + R}.$$Multiply both sides by $r_1 + r_2 + R$:

   $$r_1 + r_2 + R = 2r_1.$$Simplify:

   $$R = 2r_1 - r_1 - r_2.$$ $$R = r_1 - r_2.$$

Final Answer:

The value of $R$ is:

$$\boxed{R = r_1 - r_2}.$$