The 1968 United States presidential election in Rhode Island took place on November 5, 1968, as part of the 1968 United States presidential election. Voters chose four[2] representatives, or electors, to the Electoral College, who voted for president and vice president.
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Rhode Island was won by the Democratic candidate, Vice President Hubert Humphrey, with 64.03% of the popular vote, against the Republican candidate, former Senator and Vice President Richard Nixon, with 31.78% of the popular vote. American Independent Party candidate George Wallace also appeared on the ballot, finishing with 4.07% of the popular vote.[3][4] Despite the state trending 30 points Republican after setting a record in 1964, the state continues to overwhelmingly vote for the Democratic candidate.
Nixon was the first Republican to win the presidency without gaining the majority vote in Washington, Kent, and Newport counties.
Party | Candidate | Votes | % | |
---|---|---|---|---|
Democratic | Hubert Humphrey | 246,518 | 64.03% | |
Republican | Richard Nixon | 122,359 | 31.78% | |
American Independent | George Wallace | 15,678 | 4.07% | |
Write-in | 445 | 0.12% | ||
Total votes | 385,000 | 100% |
1968 United States presidential election in Rhode Island (by county) [5] | ||||||||
---|---|---|---|---|---|---|---|---|
County | Hubert Horatio Humphrey Democratic |
Richard Milhous Nixon Republican |
George Corley Wallace American Independent |
Various candidates Other parties | ||||
% | # | % | # | % | # | % | # | |
Providence | 67.6% | 169,246 | 28.1% | 70,320 | 4.1% | 10,374 | 0.1% | 286 |
Bristol | 59.4% | 11,561 | 38.0% | 7,403 | 2.5% | 483 | 0.1% | 28 |
Kent | 58.7% | 35,609 | 37.1% | 22,493 | 4.2% | 2,519 | 0.1% | 57 |
Newport | 58.4% | 16,251 | 37.7% | 10,504 | 3.7% | 1,043 | 0.1% | 32 |
Washington | 51.6% | 13,851 | 43.4% | 11,639 | 4.8% | 1,286 | 0.2% | 42 |