How is internal resistance of a cell related to temperature

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Internal resistance of a cell is influenced by temperature, with the relationship described by the equation r = (E - V) / I. As temperature increases, the electromotive force (E) of the cell decreases according to the Nernst Equation, which initially suggests that resistance (r) should increase. However, higher temperatures lead to increased ion dissociation in solutions, resulting in greater conductance and a decrease in resistance. It is important to note that both current (I) and terminal potential (V) change with temperature, complicating the analysis of resistance. Overall, the consensus is that internal resistance typically decreases with rising temperature due to increased ion availability.
Mr Virtual
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Relevant equations

int rest =emf - terminal potential / current
or
r=E-V/I

The attempt at a solution

Since Emf of a cell is inversely proportional to temp (From Nernst Equation of Emf of a cell), from the above equation, if temp is high, E is low, therefore r is low. So, acc. to me,
r decreases with increase in temp. But obviously, if E is decreasing, this must mean that r has increased. I am confused. Please help.

Edit: If we think in terms of solutions, then dissociation increases on heating, which means more ions available in cell, which means increase in conductance, which leads to the conclusion that resistance of the cell decreases with increase in temp.

Mr V
 
Last edited:
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I found out on the net that internal resistance decreases with increase in temperature.
 
Mr Virtual said:
Relevant equations

int rest =emf - terminal potential / current
or
r=E-V/I

it's r = (E-V)/I

and I and V do not remain constant if you change the temperature. I don;t think you can
use this formula to see what happens to R unless you measure I and V.
 

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