The emf of a galvonic cell can be calculated from:

Correct answer: D. The values of the half cell

  • A. The size of the electrode
  • B. The pH of the solution
  • C. The amount of metal in the anode
  • D. The values of the half cell

Explanation

For a galvanic cell, emf can be calculated by standard reduction potentials for two half cells. Emf is independent of pH except for the reactions involving H+ or OH- ions.Emf is independent of shape and size of electrodes,distance between electrodes and the amount of electrolyte.

This question appeared on the ETEA MDCAT 2006 paper, which you can sit online with every answer explained.

Written and checked by , M.Phil ChemistryLast updated
Report an error

The more specific you are, the faster it gets fixed. A source beats an opinion.

Prefer email? [email protected]

About Standard Hydrogen Electrode

The standard hydrogen electrode provides the reference potential of 0.00 V for measuring electrode potentials. It consists of platinum in contact with hydrogen gas at standard pressure and hydrogen ions at standard concentration, usually at 25°C, and can operate as either the oxidation or reduction half-cell.

Practise Electrochemistry

696 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Discussion

Stuck on an option, or know a faster way to get there? Ask or explain it here.

No comments yet. Be the first to explain this one.

Exams that ask Chemistry questions like this

Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.

More Standard Hydrogen Electrode questions