My physics book describes the electrochemical cell as the following diagram.

enter image description here

Let P and Q be points of the electrodes joined externally by a wire and A and B be internal points of the electrodes immersed in electrolyte. Suppose,

PA= Positive Electrode; QB= Negative Electrode

Let the potentials at all these points be V(A), V(B),V(P) and V(Q). Now,

V(P)-V(A) = V > 0

Now what I am not getting is that what will be

V(Q)-V(B) = A Positive Value Or A Negative Value???

Please Tell me With Explanation

My book says it would be negative. But Why?


Since there is no battery in the circuit I must assume that some sort of chemical reaction is moving positive charges from $B$ to $A$.

The cases to consider are that the electrodes $AP$ and $BQ$ have no resistance or the electrodes have some resistance.

If the electrodes have no resistance then $V(P) = V(A)$ and $V(Q) = V(B)$
So the potential difference across the electrodes is zero.

If the electrodes have the same resistance and given the direction of flow of the positive charges then the potential of $A$ must be greater than the potential of $P$ and the potential of $Q$ must be greater than the potential of $B$.
This assumes that no chemical reaction is going on inside the electrodes.

So $V(P)-V(A)$ is less than zero and $V(Q)-V(B)$ is greater than zero because positive charges not inside a cell flow from from a region of high potential to a region of low potential.
This is the opposite to your statement.

If you assign a value of $V$ to $V(P)-V(A)$ then $V$ will be a negative value.
Then as the resistances of the two electrodes are the same $V(Q)-V(B) = -V$ and as $V$ is the numerical value of $V$ is negative then $-V$ will be positive.

To sum up:
$V(P)-V(A)$ is negative ($P$ is at a lower potential relative to $A$) and $V(Q)-V(B)$ is positive ($Q$ is at a higher potential relative to $B$).

  • $\begingroup$ V(P)-V(A) will give voltage between those two points or simply the potential of that electrode. How a positive electrode(as mentioned in the question) can have a negative potential? Similarly for the negative electrode which has positive potential as per ur answer $\endgroup$ Oct 31 '16 at 14:37
  • $\begingroup$ $V(A) > V(P) > V(Q) > V(B)$ $\endgroup$
    – Farcher
    Oct 31 '16 at 14:46
  • $\begingroup$ there is a little bit of mistake. i have to edit my question. I am extremely sorry for this inconvenience. Really sorry $\endgroup$ Oct 31 '16 at 14:49
  • $\begingroup$ My answer is unchanged. The positive charges drop in potential at A as they go round the external circuit before arriving at B. Then a chemical reaction lifts the positive charges up in potential when moving from B to A. $\endgroup$
    – Farcher
    Oct 31 '16 at 14:57

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