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The1872 United States presidential election in Rhode Island took place on November 5, 1872. All contemporary 37 states were part of the1872 United States presidential election. The state voters chose four electors to theElectoral College, which selected thepresident andvice president.
Rhode Island was won by theRepublican nominees, incumbentPresidentUlysses S. Grant ofIllinois and his running mateSenatorHenry Wilson ofMassachusetts. Grant and Wilson defeated theLiberal Republican andDemocratic nominees, formerCongressmanHorace Greeley ofNew York and his running mate formerSenator andGovernorBenjamin Gratz Brown ofMissouri by a margin of 43.88%.
With 71.94% of the popular vote, Rhode Island would be Grant's third strongest victory in terms of percentage in the popular vote afterVermont andSouth Carolina.[1] It remains the best Republican performance in Rhode Island's history and the second-best by any candidate since theformation of the Democratic party in 1828 afterLyndon B. Johnson’s 80.87% in1964.[2]
| 1872 United States presidential election in Rhode Island[3] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | Ulysses S. Grant ofIllinois | Henry Wilson ofMassachusetts | 13,665 | 71.94% | 4 | 100.00% | ||
| Liberal Republican | Horace Greeley ofNew York | Benjamin Gratz Brown ofMissouri | 5,329 | 28.06% | 0 | 0.00% | ||
| Total | 18,994 | 100.00% | 4 | 100.00% | ||||
State and district results of the1872 United States presidential election | ||
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