Number of electron added on both sides of oxidation and reduction half reactions are balanced:
Correct answer: D. Before Adding two half reactions
- A. At the start of the procedure
- B. Somewhere in the middle of balancing
- C. After Adding two half reactions
- D. Before Adding two half reactions
Explanation
In electrochemistry, it is crucial to balance the number of electrons lost in the oxidation half-reaction with the number of electrons gained in the reduction half-reaction before adding the two half-reactions together. This ensures that the overall redox reaction is balanced in terms of both mass and charge. Therefore, the correct step to balance electrons is before adding the two half reactions. The other options are incorrect because balancing the electrons at any other stage would not result in a properly balanced equation.
Last updated
About Electrochemistry
Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
Practise Electrochemistry
696 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?
+3 and +5 oxidation states are often shown by:
3 faraday of electricity is passed through molten Al2O3 , aqueous solution of CuSO4 and molten NaCl taken in three different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in the ratio of:
5.3 gm of M2CO3 is dissolved in 150 ml of 1N HCl. The unused acid required 100 ml of 0.5N NaOH. Hence equivalent weight of M is:
8H++ MnO4- → Mn2+ + 4H2O , which one is correct about given equation?