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1888 United States presidential election in Rhode Island

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Main article:1888 United States presidential election
1888 United States presidential election in Rhode Island

← 1884
November 6, 1888
1892 →
 
NomineeBenjamin HarrisonGrover Cleveland
PartyRepublicanDemocratic
Home stateIndianaNew York
Running mateLevi P. MortonAllen G. Thurman
Electoral vote40
Popular vote21,96917,530
Percentage53.88%42.99%

County Results
Municipality Results

Harrison

  40–50%
  50–60%
  60–70%
  70–80%
  80–90%

Cleveland

  40–50%
  50–60%


President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

Elections in Rhode Island
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The1888 United States presidential election in Rhode Island took place on November 6, 1888, as part of the1888 United States presidential election. Voters chose four representatives, or electors to theElectoral College, who voted forpresident andvice president.

Rhode Island voted for theRepublican nominee,Benjamin Harrison, over theDemocratic nominee, incumbentPresidentGrover Cleveland. Harrison won the state by a margin of 10.89%.

Results

[edit]
1888 United States presidential election in Rhode Island[1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
RepublicanBenjamin Harrison ofIndianaLevi Parsons Morton ofNew York21,96953.88%4100.00%
DemocraticGrover Cleveland ofNew York (incumbent)Allen Granberry Thurman ofOhio17,53042.99%00.00%
ProhibitionClinton Bowen Fisk ofNew JerseyJohn Anderson Brooks ofMissouri1,2513.07%00.00%
Union LaborAlson Streeter ofIllinoisCharles E. Cunningham ofArkansas180.04%00.00%
N/AOthersOthers70.02%00.00%
Total40,775100.00%4100.00%

See also

[edit]

References

[edit]
  1. ^"1888 Presidential General Election Results - Rhode Island". U.S. Election Atlas. RetrievedDecember 23, 2013.
State and district results of the1888 United States presidential election
Electoral map, 1888 election


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