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It could possibly be beneficial if the experiment is successful.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

The reaction 2kalsi3o8 + 2(h+ + hco3-) + h2o  →  al2si2o5(oh)4 + 2k+ + 2hco3- + 4sio2 shows that feldspar kalsi3o8 is chemically altered by hydrolysis to form chalky-looking clay mineral kaolinite al2si2o5(oh)4. the chemical breakdown of potassium feldspar due to exposure to acidic water results to the mineral binding with the water molecule to become kaolinite with silica and potassium and bicarbonate ions in solution.

Ch3cooh i think this is the answer