[MikeK]

Citric acid will react directly with the tin in the babbitt alloy matrix, forming soluble stannous citrate. (Yes, it will eat the babbitt).

Of greater concern is what happens to the rusty block. The first reaction goes like this:
C6H8O7 + Fe2O3——> 2 FeO + 6 CO + 2 H2O + 2 H2
Citric acid + red rust yields gray iron oxide, carbon monoxide, water and hydrogen.

Notice you will be producing copious quantities of carbon monoxide! Of equal concern is the reactions that follow. Upon removal of the 'rust' and flaking of the gray iron oxide you will have bare iron exposed that WILL react, forming soluble ferrous citrate. (It will eat into the bare metal)

Next, with both the babbitt and iron touching, the citric acid solution will become a battery. The electro potential difference between the tin and the iron will cause the tin to react and go into solution. (bye bye babbitt)

Summing it all, after the citric acid solution has first eaten so deeply into the babbitt alloy matrix that the tin is no longer chemically available the excess citric acid will begin working on the iron and will not stop until one of the two reactants is consumed.

Citric acid may work on lightly, evenly rusted tools if you watch closely and stop, but I would not use it on ANY part made of dissimilar metals/alloys that are bonded together, nor any part with light/heavy rust variations or pits.