<p>Which of the following represents a process in which a species is reduced?</p>
<p>(a) Ca(s) → Ca 2+(aq)
(b) Hg(l) → Hg2 2+(aq)
(c) Fe 2+(aq) → Fe 3+(aq)
(d) NO3 (aq) → NO(g)
(e) SO3 2(aq) → SO4 2(aq)</p>
<p>LEO goes GER</p>
<p>For it to be reduced it needs to gain electrons, but none of these things are gaining electrons...</p>
<p>Cd2+(aq) + 2 e– →← Cd(s) Eo = -.41 V
Cu+(aq) + e– ←→ Cu(s) Eo = +.52 V
Ag+(aq) + e– ←→ Ag(s) Eo = +.80 V</p>
<p>Eo = –0 .41 V Eo = +0 .52 V Eo = +0 .80 V
15 . Based on the standard electrode potentials given above, which of the following is the strongest reducing agent?</p>
<p>(a) Cd(s) (b) Cd2+(aq) (c) Cu(s) (d) Ag(s) (e) Ag+(aq)</p>
<p>Please explain!</p>
<p>1st question:
First of all, gaining electrons is not only the definition of reduction. Losing of oxygen atoms too should be kept in concern and above all decrease in oxidation number is also known as reduction.
Having said that, if we look at all the options,
only D,in which the atom N, where the number of oxygen atoms is decreasing, is undergoing a reduction…</p>
<p>Thanks sammy, I wasn’t aware of that alternate definition, but what about #2?</p>
<p>For number 2 you should know that the reactions given are for standard electrode potentials and they are the indicators of how easily the reactant (L.H.S) is reduced to the products(R.H.S).Also if one atom reduces then it oxidizes other one. So the less positive the Eo, the difficult it is to reduce and the less oxidising capacity and more reducing capacity it has.
By looking at these facts, I think the correct answer is D Ag(S)…</p>
<p>^Thats what I thought too, but the answer is (A). This is from the collegeboard.</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Here is where I got the question from.</p>
<p>The answer is A because the strongest reducing agent will have the highest oxidation potential and the ones given are standard [reduction] potentials. Flip the sign for oxidation potentials and the highest one is Cadmium’s</p>
<p>By flipping the signs the highest one is +.41 V. Wouldn’t that be Cd 2+, or (B)? Why is it Cd?</p>
<p>Lol, 2+ is the highest oxidation state of cadmium. Since Cd 2+ is already is in a oxidation state of 2+, it cannot be further oxidized. So the answer is the metal, Cd.</p>
<p>Reducing agent= thing getting oxidized.</p>